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Mar 12 at 22:44 history edited GH from MO
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Sep 19, 2022 at 11:45 comment added Jeppe Stig Nielsen That OEIS entry was updated now.
Sep 8, 2022 at 14:49 comment added Jeppe Stig Nielsen Procrastination comment. If I fix $a=2$, I find (with a small PARI/GP function) "good" $b$ values 909, 4995825 (your example), 28212939. Searching OEIS gives A245510; they do not have better examples.
Sep 8, 2022 at 13:48 comment added Roland Bacher Thanks. I was indeed aware of this: $b$ (the basis of powers) tends to have many small prime-divisors. Your observation implies that $b$ is a perfect square if it is fixed and $n$ can be arbitrarily large.
Sep 8, 2022 at 9:18 comment added Gerry Myerson If you have a prime $p$ and a set $\{a_1,a_2,\dots,a_n\}$ of positive integers which covers all congruence classes modulo $p$, then you can't have $m\ge p$ such that $m+a_i$, $1\le i\le n$, are all prime. E.g., $m+2$, $m+4$, $m+6$ can't all be prime (for $m>1$) since one of them will be a multiple of three, but $m+2$, $m+4$, $m+8$ can all be prime and, conjecturally, are all prime for infinitely many $m$. We say $\{2,4,8\}$ is admissible, $\{2,4,6\}$ isn't. en.wikipedia.org/wiki/Prime_k-tuple#Admissibility
Sep 8, 2022 at 7:43 comment added Roland Bacher Thanks Gerry. I do not understand the link between admissibility and primality: Is admissibility a necessary prerequisites for the conjectures?
Sep 8, 2022 at 5:17 comment added Gerry Myerson There is no prime $p$ such that $4^k$, $k=1,2,\dots,n$ covers all the residue classes mod $p$ (since $4$ is a quadratic residue mod $p$, its powers can only be quadratic residues), so those powers of four constitute an admissible set. Then standard (but unproved) conjectures imply that for each $n$ there exist $a$ such that $a+4^k$ is prime for $1\le k\le n$.
Sep 7, 2022 at 17:59 history edited Michael Hardy CC BY-SA 4.0
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Sep 7, 2022 at 17:40 history asked Roland Bacher CC BY-SA 4.0